x² - 4x + 4 = 0 

Step 1: Points to be plotted: (-4,36), (-3, 25), (-2, 16), (-1, 9), (0, 4), (1, 1), (2, 0), (3, 1), (4,4)
Step 2: The point of intersection of the curve with x axis is (2, 0)
Step 3:
 
Since there is only one point of intersection with x axis, the quadratic equation X² - 4x + 4 = 0 has real and equal roots.
∴ Solution{2,2}

Step 1 : Draw the graph of y = x² + x - 2 by preparing the table of values as below

Step 2 : To solve x² + x - 2 = 0 subtract x² + x - 2 = 0 from y = x² + x - 2

The equation y = 0 represents the X axis.
Step 3 : Mark the point of intersection of the curve x² + x - 2 with the X axis. That is (–2,0) and (1,0)
Step 4 : The x coordinates of the respective points form the solution set {−2,1} for x² + x - 2 = 0

Given, radius = 4 cm
Construction
Step 1 : With O as the centre, draw a circle of radius 4 cm.
Step 2 : Take a point L on the circle. Through L draw any chord LM.
Step 3 : Take a point M distinct from L and N on the circle, so that L, M and N are in anti clockwise direction. Join LN and NM.
Step 4 : Through L draw a tangent TT' such that \(\angle\)TLM =\(\angle\)MNL
Step 5 : TT' is the required tangent.

Construction:
Steps (1) Draw a line segment BC = 5.6 cm
Steps (2) At B, draw BE such that \(\angle CBE={ 60 }^{ 0 }\)
Steps (3) At B draw BF such that \(\angle EBF={ 90 }^{ 0 }\)
Steps (4) Drawn the perpendicular bisector to BC, which intersects BF at O and BC at G.
Steps (5) With O as centre and OB as radius draw a circle
Steps (6) From B, marked an arc of 4 cm on BC at D.
Steps (7) The perpendicular bisector intersects the circle at I. Joined ID.
Steps (8) ID produced meets the circle at A. Now joined AB and AC. Then \(\triangle\)ABC is the required triangle.